Axial Loading

 Internal Force–Deformation–Displacement

Members in simple axial loading are either stretched or compressed in their long direction. The forces at the ends can be related to the displacements of points on the member.

•  A cord that elongates due to tensile forces at its ends as in musical instrument is an example of axial loading.
• A long straight member,which could withstand more complex loadings, is in axial loading if the forces act along its length.
  
  Cords or cables can only be in equilibrium if the forces act along their lengths and they produce tension.
• Find the internal force P from equilibrium of a portion of the body. Stress and deformation are determined by forces acting internally.
          For this simple case, the internal force P is equal in magnitude to F0. So if the force on each end is F₀=100 N, the internal force is P 100 N (not 200 N).
  In more complex situations in which several external forces act, one must take care in relating internal and external forces.
• Combine relations between force, stress, strain, and elongation to relate the internal force to the elongation.
  As we  know, the stress sand strain  are uniform in a bar of uniform material (elastic modulus E) and cross-section (area A) that is subjected to axial forces. We studied these relations:
  

  Stress is related to internal force	Strain is related to elongation and bar length L	Stress is related to strain if the material is elastic
  σ = P/A	ε = δ/L	E = σ/ε

  Combine these equations to relate the elongation, d, to the internal force, P :   δ=εL=σL/E =PL/EA
  
  δ=PL/EA                 or          P=δEA/L
•  A bar under axial loading is analogous to an elastic spring.
       The force-elongation relation of a bar,P=EAδ/A , is analogous to the spring law studied in physics:
                 F=kx                                       F=force, k=spring constant or stiffness, x=displacement
  EA/L is analogous to the spring constant k. We call EA/L the axial stiffness of a bar of length L.
  We apply     P=(EA/L)δ  carefully:      P= internal force in a segment, not the external force
                                                             δ=elongation of a segment, not the displacement of one point.
  To use this relation, E, A, and P must be uniform over the segment of length L
• Displacement of a point on a body-Individual points on a body move possibly by different amounts in response to forces. We define the displacement u as the horizontal distance moved by a point. We define v as the vertical displacement. The sign of u signifies whether the displacement is left or right (v, up or down).
        A bar with end points A and B is shown first in its unloaded configuration. The bar is uniformly strained by equal and opposite forces (not drawn). Displacements of the end points are given in the table at the right. Consider the relation between u at the two ends and the change in length, δ.Here choose the signs to mean: u > 0 to right, u < 0 to left,δ > 0 elongates,δ < 0 contracts.
  
        One can see that the displacement at one point and δ dare not uniquely related to each other. Instead,δ is equal to the difference in u at the two end points. For the signs used above:δ=u_B-u_A 
        We choose the meanings of the signs for displacement and elongation as convenient for a particular problem. Depending on the signs, the relation δ = u_B - u_A     or  δ=u_A - u_B will be correct. 
• Use the relations between elongation and displacements to find displacement at any point along a bar.
• Use the same relation between dand uto find the displacement at any point along the length of a
  uniformly strained bar. Given displacements of points Aand B, what is the displacement of point C?
  Strain is uniform:           ε = δ_AB/L = (u_B-u_A)/L = [3-(-1)]/8=4/8=0.5
  Elongation of AC:         δ_AC = L_AC(ε) = (3)(0.5)=1.5
  Displacement of C:       δ_AC = u_C-u_A ⇒ u_c = δ_AC + u_A = 1.5 + (-1) = 0.5

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Varying Internal Force

So far, the body has been loaded axially with two equal and opposite forces at its ends. We now consider situations in which a bar is loaded axially at more than two points.Several axial forces acting at different points along a member in an applications.

• A member in axial loading may not simply be loaded by one force at each end.Rather, there can be loads acting at intermediate points along the length.For example, this vertical post has axial loads due to the weight of each floor beam,and a support reaction from the ground.
  
  
• When multiple axial forces act on a body, the deformation can be different between each successive pair of forces.
  Hands that grip this bungee cord at points A, C, and B apply forces that deform the cord.  
  To illustrate the deformation of a cord, we have drawn lines that divide the cord into segments that are initially of equal length. The five segments between A and C elongate equally, and the three segments between C and Belong ate equally. Between A and C, the strain, stress, and internal force are uniform, and they differ from the uniform values between C and B. 
• Find the internal force at any cross-section by imposing equilibrium on an FBD that includes a portion of the body ending at the cross-section.
           Each portion of the cord must be in equilibrium. We can relate the internal forces to the external forces by imposing equilibrium on a portion of the cord.
          Think of the cord as composed of two portions: from A to some point between A and C and from that point to B. The two portions exert equal and opposite internal forces on each other where they join. For example, PAC denotes the internal force between two portions that meet at a cross-section between A and C.
           
                 ΣF_x = -1 + P_AC = 0               or              ΣF_x = - P_AC - 3 + 4 = 0
  
  
      Now think of the two portions joining at some point between C and B.
          
               ΣF_x = -1 - 3 + P_CB = 0               or              ΣF_x = - P_CB + 4 = 0
  The internal force,P_AC = 1 N,is uniform from A to C, and P_CB N is uniform from B to C.
•  Because the internal force is uniform, but different in each of the segments AC and CB,we calculate the stress and elongation for each of those two segments separately.
  Since the internal force, area, and material are uniform within the segments AC and CB, we can apply relations for uniformly axially loaded bodies to each separately:
• Combine successive relations between external and internal forces, stress, strain, elongation, and displacement as needed.
        When there are several axial forces along the length, our goal is often to find stress and displacements everywhere. The basic formulas that relate internal force, stress, strain, and elongation are applicable when P,A,and E are constant along a segment.
• Recognize tension vs.compression, and use the sign of P to distinguish them.

  Tension: all forces in outward sense on face Compression: all forces in outward sense on face

  Use P > 0 for tension and P < 0 for compression, e.g., P = -2 kN means 2 kN in compression.The direction of the internal force in a segment may be hard to guess in advance.To simplify our bookkeeping, we usually draw the senses of unknown internal forces assuming they are in tension. When we eventually solve, if P > 0, the internal force is tensile, and if P < 0 , the internal force is compressive. 
• A bar with multiple forces is analyzed by decomposing it into segments in which P, E,and A are uniform.
  To illustrate the procedure: consider one bar with a change in cross-section that is attached rigidly to a base. A bar of different material is screwed into the first. External forces are applied to the bars at the points shown.
  Identify segments in which internal force P, E, and A are all constant.
  Do this by recognizing and labeling points where something changes
  

  Point	B	C	D	E
  Change	Int.Force	Modulus	Area	Int.Force

  Label segments as 1, 2, 3, 4, and 5 (could use AB, BC, CD, DE and EF instead). For each segment, find P from an FBD extending from one end to a cross-section in that segment.Unknown P is drawn so it is in tension, if positive.
  
  
             ⇒ P₁ = 1KN , P₂ = P₃ = P₄ = 2KN , P₅ = -4 KN
  Compute the stretch of each segment using P, L, E, and A appropriate to each:
  Stretches:
  Relate displacements (take v 0 upward) to stretches
  δ₅ = ν_E - ν_F , ν_F = 0 ⇒ ν_E = δ₅,
  δ₄ = ν_D - ν_E  ⇒ ν_D = δ₄ + ν_E =δ₄ + δ₅ , and so forth.
  

• Continuously varying bar. Integrate to find elongation when P, E, or A vary continuously along the bar.
  The area or modulus may vary continuously along a bar (with x). Also, when external forces are applied in a distributed manner, the internal force varies with x : P(x). We assume the same relation between force, stress, and elongation still applies to a slice of length dx. So the elongation of a segment from cross-section, xA, to another, xB, is found from integration:
  
  •  σ(x) = P(x)/A(x)         
  •  dδ = P(x)dx/E(x)A(x)             
  •  δ = ∫ P(x)dx / E(x)A(x) 

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